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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ω and 0.1 Ω respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.
Correct answer is between '9.5,12'. Can you explain this answer?
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A 220 V DC series motor runs drawing a current of 30 A from the supply...
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A 220 V DC series motor runs drawing a current of 30 A from the supply...
Solution:
Given parameters:
- Supply voltage (V) = 220 V
- Armature circuit resistance (Ra) = 0.4 Ω
- Field circuit resistance (Rf) = 0.1 Ω
- Armature current (Ia) = 30 A
- Flux (Φ) ∝ Ia
- Load torque (T) ∝ N^2 (where N is the speed)
We know that the speed of the DC series motor is given by:
N = (V - IaRa) / (ΦK)
where K is a constant.
We can write the torque equation as:
T = (ΦIa / K) * N
or
T = (ΦIa / K) * [(V - IaRa) / (ΦK)]
or
T = (ΦIa / K^2) * (V - IaRa)
Since Φ ∝ Ia, we can write:
T ∝ Ia^2 * (V - IaRa)
If we reduce the speed of the motor by 50%, the new speed (N') will be:
N' = N / 2
or
N' = (V - IaRa + RextIa) / (Φ'K)
where Φ' is the new flux and Rext is the resistance that should be added in series with the armature.
We can write the torque equation for the new speed as:
T' = (Φ'Ia / K) * N'
or
T' = (Φ'Ia / K^2) * (V - IaRa + RextIa) / 2
or
T' = (Φ'Ia / K^2) * [(V - IaRa) / 2 + RextIa / 2]
Since Φ' ∝ Ia, we can write:
T' ∝ Ia^2 * [(V - IaRa) / 2 + RextIa / 2]
To reduce the speed by 50%, we need to reduce the torque by 75% (since T' = 0.25T).
Thus, we can write:
0.25T = Ia^2 * [(V - IaRa) / 2 + RextIa / 2]
or
0.25 * Ia^2 * (V - IaRa) = Ia^2 * Rext / 2
or
Rext = 0.5 * (V - IaRa) / 0.25 = 2 * (V - IaRa)
Substituting the given values, we get:
Rext = 2 * (220 - 30 * 0.4) = 2 * 208 = 416 Ω
Therefore, the resistance in ohms that should be added in series with the armature to reduce the speed of the motor by 50% is 416 Ω, which lies between 9.5 Ω and 12 Ω.
Given parameters:
- Supply voltage (V) = 220 V
- Armature circuit resistance (Ra) = 0.4 Ω
- Field circuit resistance (Rf) = 0.1 Ω
- Armature current (Ia) = 30 A
- Flux (Φ) ∝ Ia
- Load torque (T) ∝ N^2 (where N is the speed)
We know that the speed of the DC series motor is given by:
N = (V - IaRa) / (ΦK)
where K is a constant.
We can write the torque equation as:
T = (ΦIa / K) * N
or
T = (ΦIa / K) * [(V - IaRa) / (ΦK)]
or
T = (ΦIa / K^2) * (V - IaRa)
Since Φ ∝ Ia, we can write:
T ∝ Ia^2 * (V - IaRa)
If we reduce the speed of the motor by 50%, the new speed (N') will be:
N' = N / 2
or
N' = (V - IaRa + RextIa) / (Φ'K)
where Φ' is the new flux and Rext is the resistance that should be added in series with the armature.
We can write the torque equation for the new speed as:
T' = (Φ'Ia / K) * N'
or
T' = (Φ'Ia / K^2) * (V - IaRa + RextIa) / 2
or
T' = (Φ'Ia / K^2) * [(V - IaRa) / 2 + RextIa / 2]
Since Φ' ∝ Ia, we can write:
T' ∝ Ia^2 * [(V - IaRa) / 2 + RextIa / 2]
To reduce the speed by 50%, we need to reduce the torque by 75% (since T' = 0.25T).
Thus, we can write:
0.25T = Ia^2 * [(V - IaRa) / 2 + RextIa / 2]
or
0.25 * Ia^2 * (V - IaRa) = Ia^2 * Rext / 2
or
Rext = 0.5 * (V - IaRa) / 0.25 = 2 * (V - IaRa)
Substituting the given values, we get:
Rext = 2 * (220 - 30 * 0.4) = 2 * 208 = 416 Ω
Therefore, the resistance in ohms that should be added in series with the armature to reduce the speed of the motor by 50% is 416 Ω, which lies between 9.5 Ω and 12 Ω.
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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?
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A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?.
A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4 Ωand 0.1 Ωrespectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.Correct answer is between '9.5,12'. Can you explain this answer?.
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